Consider the function f(x)=4sqrt(x)+4 on the interval [2,5] . Find the average or mean slope of the function on this interval _______ <---A


By the Mean Value Theorem, we know there exists a c in the open interval (2,5) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it. _______ <---B

A=?
B=?

average slope=(f(5)-f(2))/(5-2)

set average slope above to equal
f=-2/sqrt(x) then solve for x.

To find the average or mean slope of the function f(x) on the interval [2,5], we can use the formula:

Mean Slope = (f(5) - f(2)) / (5 - 2)

First, let's find f(5):
f(x) = 4√x + 4
f(5) = 4√5 + 4

Next, let's find f(2):
f(2) = 4√2 + 4

Now, we can substitute these values into the formula for mean slope:
Mean Slope = (4√5 + 4 - 4√2 - 4) / (5 - 2)

Simplifying this expression gives us the value of the average or mean slope on the interval [2,5].

So, A = (4√5 + 4 - 4√2 - 4) / 3

To find the value of c mentioned in the Mean Value Theorem, we need to find f'(x) which is the derivative of f(x).

Taking the derivative of f(x) = 4√x + 4:
f'(x) = (4/x^(-1/2))/2 = 2/x^(1/2) = 2√x

Now, we need to find the value of c where f'(c) is equal to the mean slope.

So, B is the solution to the equation 2√c = (4√5 + 4 - 4√2 - 4) / 3. We need to solve for c by isolating it.